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(D) The given circuit is an infinite ladder network. Due to the symmetry of the circuit,the potential difference across the vertical $1 \mu F$ capacit

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$\frac{4}{3} \mu F$

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(D) The given circuit is an infinite ladder network. Due to the symmetry of the circuit,the potential difference across the vertical $1 \mu F$ capacitors is zero. Thus,these capacitors can be removed from the circuit.After removing the vertical capacitors,the circuit simplifies into two parallel branches,each containing an infinite series of capacitors with values $1 \mu F, 3 \mu F, 9 \mu F, 27 \mu F, \dots$ in a geometric progression.Let the equivalent capacitance of one branch be $C'$. The reciprocal of the equivalent capacitance for one branch is given by the sum of the reciprocals of the individual capacitors:$\frac{1}{C'} = \frac{1}{1} + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$This is an infinite geometric series with the first term $a = 1$ and common ratio $r = \frac{1}{3}$.The sum of an infinite geometric series is $S = \frac{a}{1-r}$.$\frac{1}{C'} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \mu F^{-1}$.Therefore,$C' = \frac{2}{3} \mu F$.Since there are two such branches connected in parallel between $A$ and $B$,the total equivalent capacitance $C_{AB}$ is:$C_{AB} = C' + C' = 2 	imes \frac{2}{3} \mu F = \frac{4}{3} \mu F$.

The equivalent capacitance between $A$ and $B$ in the given figure is

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